When a polynomial is divided by (x+2), the remainder is -19. When the same polynomial is divided by (x-1), the remainder is 2, how do you determine the remainder when the polynomial is divided by (x+2)(x-1)?

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Jim H Share
Jul 2, 2015

Answer:

The remainder is #7x-5#.

Explanation:

Call the polynomial #P(x)#.

Because #P(x)# divided by #x+2# leaves a remainder of #-19#, we know that:
#P(x) = (x+2)Q_1(x) - 19#. (Division Algorithm)

Because #P(x)# divided by #x-1# leaves a remainder of #2#, we know that #P(1) = 2# (Remainder Theorem)

From these two facts we get:

#P(1) = (1+2)Q_1(1)-19 = 2#

So #3Q_1(1) = 21#

and #Q_1(1) = 7#

Applying the Division Algorithm and the Remainder Theorem to #Q_1(x)# we learn that:

#Q_1(x) = (x-1)Q_2(x)+7#

Subtitutuing in #P(x) = (x+2)Q_1(x) - 19#, we get:

#P(x) = (x+2)[(x-1)Q_2(x)+7] - 19#

# = (x+2)(x-1)Q_2(x) + 7(x+2) - 19#

# = (x+2)(x-1)Q_2(x) + 7x+ 14 - 19#

# = (x+2)(x-1)Q_2(x) + [7x - 5]#

The remainder is #7x-5#.

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9
May 28, 2017

Answer:

#7x-5.#

Explanation:

Let the Poly. in question be #p(x).#

We know that, the Degree of the Remainder Poly. is strictly

less than that of the Divisor Poly.

So, when #p(x)# is diveded by a Quadr. Poly. #(x+2)(x-1)#, the

Remainder Ploly. must have a degree #1, or, 0.#

Therefore, let us suppose that, the desired remainder is, #ax+b.#

Symbolically, this can be said, as, let,

#p(x)=(x+2)(x-1)q(x)+(ax+b)..................(ast).#

Now, #p(x),# when divided by #(x+2)# leaves remainder #-19,#

# rArr p(-2)=-19.#

#rArr -2a+b=-19,............[because, (ast)]...........(ast_1).#

On the same lines, #p(1)=2 rArr a+b=2.............(ast_2).#

Solving #(ast_1) and (ast_2), a=7, b=-5,# giving the desired

remainder, #7x-5,# as Respected JIM H. and George C. have

readily obtained.

Enjoy Maths.!

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4
May 28, 2017

Answer:

The remainder when dividing #f(x)# by #(x+2)(x-1)# is #(7x-5)#

Explanation:

Note that #(x+2)(x-1) = x^2+x-2# is quadratic, so if we divide another polynomial by it then the remainder will be #0#, a non-zero constant or a linear polynomial. Any remainder of greater degree could be divided further.

Suppose #f(x) = (x+2)(x-1)k(x) + rx + s#

Then:

#f(x)/(x+2) = (x-1)k(x) + (rx+s)/(x+2)#

#=(x-1)k(x) + (rx + 2r -2r + s)/(x+2)#

#=(x-1)k(x) + r + (s-2r)/(x+2)#

Hence [1]: #s-2r = -19#

#f(x)/(x-1) = (x+2)k(x) + (rx+s)/(x-1)#

#=(x+2)k(x) + (rx-r+r+s)/(x-1)#

#=(x+2)k(x)+r+(r+s)/(x-1)#

Hence [2]: #r+s = 2#

Subtract [2] from [1] to get #-3r = -21#, hence #r = 7#

Then #s = 2 - r = -5#

So the remainder when dividing #f(x)# by #(x+2)(x-1)# is #7x-5#

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